Answer to Question #157273 in Statistics and Probability for Vincent Miyato

Question

A survey commissioned to assess whether people in some country would welcome the constitution of a commission of inquiry into the Privatization process of that country which took place two decades ago reported that 51% of the respondents felt it was total wastage of the already constrained country's resources. Of the respondents who were age 45 or older, 70% believed the commission of inquiry was necessary. Of the people surveyed,57% were under age of 45. One respondent is selected randomly.

(i) What is the probability that the person is younger than age 45 and believes that the commission of inquiry is necessary?

(ii) If the person selected believes that the commission of inquiry is necessary, what is the probability that the person is 45 years old or older?

(iii)What is the probability that the person is younger than age 45 or believes the commission of inquiry is necessary?

Answer 1

(a) Let $A$ and $B$ be two events such that , $A=$ person is younger than age $45$

$B=$ person believe that the commission inquiry is necessary

Then $P(A)=\frac{57}{100}$ and $P(B)=\frac{49}{100}$

Therefore the required probability is = $P(A\cap B)= P(A).P(B)=\frac {57}{100} .\frac{49}{100}=0.2793$

(b) Let $E_1,E_2$ and $B$ be three events such that, $E_1=$ person is age $45$ or older

$E_2=$ person is younger than $45$

$B=$ person believe that the commission inquiry is necessary

Then $P(E_1)=\frac{43}{100}$ , $P(E_2)=\frac{57}{100}$ and $P(B/E_1)=\frac{70}{100}$, $P(B/E_2)=\frac{49}{100}$

Now we have to find $P(E_1/B).$

According to Bay's theorem, $P(E_1/B)=\frac{P(E_1).P(B/E_1)}{P(E_1).P(B/E_1)+P(E_2).P(B/E_2)}$

$=\frac{\frac{43}{100}×\frac{70}{100}}{(\frac{43}{100}×\frac{70}{100})+(\frac{57}{100}×\frac{49}{100})}$

$=\frac{3010}{5803}$

(c) Let $A$ and $B$ be two events such that , $A=$ person is younger than age $45$

$B=$ person believe that the commission inquiry is necessary

Then $P(A)=\frac{57}{100}$ and $P(B)=\frac{49}{100}$

Therefore the required probability is $P(A\cup B)$

Now $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=P(A)+P(B)-P(A).P(B)$

$=\frac{57}{100}+\frac{49}{100}-\frac{57}{100}.\frac{49}{100}$

$=\frac{7807}{10000}=0.7807$